Question 6 Exercise 4.1

Solutions of Question 6 of Exercise 4.1 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

The general recursive definition formula defined for Pascal sequences is $$P_0=1, P_{r+1}=\dfrac{n-r}{r+1} P_r, \text{ where } r=0,1,2,3,\ldots.$$

Find the Pascal sequence for $n=5$ by using its general recursive definition.

For $n=5$, we have Pascal sequence as follows: $$P_0=1, P_{r+1}=\dfrac{5-r}{r+1} P_r, \text{ where } r=0,1,2,3,\ldots.$$ For $r=0$ $$\begin{align}&P_{0+1}=\dfrac{5-0}{0+1} P_0\\ \implies &P_1=5.\end{align}$$ For $r=1$ $$\begin{align}&P_{1+1}=\dfrac{5-1}{1+1} P_1\\ \implies &P_2=2\cdot 5=10.\end{align}$$ For $r=2$ $$\begin{align}&P_{2+1}=\dfrac{5-2}{2+1} P_2\\ \implies &P_3=1\cdot 10=10\end{align}$$ For $r=3$ $$\begin{align}&P_{3+1}=\dfrac{5-3}{3+1} P_3\\ \implies &P_4=\dfrac{10}{2}=5.\end{align}$$ For $r=4$ $$\begin{align}&P_{4+1}=\dfrac{5-4}{4+1} P_4\\ \implies &P_5=\dfrac{5}{5}=1\end{align}$$ For $r=5$ $$\begin{align}&P_{5+1}=\dfrac{5-5}{5+1} P_4\\ \implies &P_6=0\end{align}$$ So, $0=P_7=P_8=...$.

Hence the Pascal sequence for $n=5$ is $1,5,10,10,5,1,0,0,0, \ldots$.

Find the Pascal sequence for $n=6$ by using its general recursive definition.

As we know the general definition of Pascal sequence is $$P_0=1, P_{r+1}=\dfrac{n-r}{r+1} P_r, \text{ where } r=0,1,2,3,\ldots.$$ When $n=6$, then $$P_0=1, P_{r+1}=\dfrac{6-r}{r+1} P_r, \text{ where } r=0,1,2,3,\ldots.$$ Now for $r=0$ $$\begin{align}&P_{0+1}=\dfrac{6-0}{0+1} P_0\\ \implies &P_1=6\end{align}$$ For $r=1$ $$\begin{align}&P_{1+1} = \dfrac{6-1}{1+1} P_1\\ \implies & P_2 = \dfrac{5}{2} \cdot 6=15\end{align}$$ For $r=2$ $$\begin{align}&P_{2+1}=\dfrac{6-2}{2+1} P_2\\ \implies &P_3=\dfrac{4}{3} \cdot 15=20\end{align}$$ For $r=3$ $$\begin{align}&P_{3+1}=\dfrac{6-3}{3+1} P_3\\ \implies &P_4=\dfrac{3}{4} \cdot 20=15\end{align}$$ For $r=4$ $$\begin{align}&P_{4+1}=\dfrac{6-4}{4+1} P_4\\ \implies &P_5=\dfrac{2}{5} \cdot 15=6\end{align}$$ For $r=5$ $$\begin{align}&P_{5+1}=\dfrac{6-5}{5+1} P_5\\ \implies &P_{6}=\dfrac{1}{6} \cdot 6=1\end{align}$$ For $r=6$ $$\begin{align}&P_{6+1}=\dfrac{6-6}{6+1} P_6\\ \implies &P_{7}=0\end{align}$$ So $0=P_8=P_9=...$,

hence the Pascal sequence of $n=6$ is $1,6,15,20,15,6,1,0,0,0,\ldots$.

Find the Pascal sequence for $n=8$ by using its general recursive definition.

As we know the general definition of Pascal sequence is $$P_0=1, P_{r+1}=\dfrac{n-r}{r+1} P_r, \text{ where } r=0,1,2,3,\ldots.$$ When $n=8$, then $$P_0=1, P_{r+1}=\dfrac{8-r}{r+1} P_r, \text{ where } r=0,1,2,3,\ldots.$$ For $r=0$ $$\begin{align}&P_{0+1}=\dfrac{8-0}{0+1} P_0\\ \implies &P_1=8\end{align}$$ For $r=1$ $$\begin{align}&P_{1+1}=\dfrac{8-1}{1+1} P_1\\ \implies &P_2=\dfrac{7}{2} \cdot 8=28\end{align}$$ For $r=2$ $$\begin{align}&P_{2+1}=\dfrac{8-2}{2+1} P_2\\ \implies &P_3=\dfrac{6}{3} \cdot 28=56\end{align}$$ For $r=3$ $$\begin{align}&P_{3+1}=\dfrac{8-3}{3+1} P_3\\ \implies &P_4=\dfrac{5}{4} \cdot 56=70\end{align}$$ For $r=4$ $$\begin{align}&P_{4+1}=\dfrac{8-4}{4-1} P_4\\ \implies &P_5=\dfrac{4}{5} \cdot 70=56\end{align}$$ For $r=5$ $$\begin{align}&P_{5+1}=\dfrac{8-5}{5-1} P_5\\ \implies &P_6=\dfrac{3}{6} \cdot 56=28\end{align}$$ For $r=6$ $$\begin{align}&P_{6+1}=\dfrac{8-6}{6+1} P_6\\ \implies &P_7=\dfrac{2}{7} \cdot 28=8\end{align}$$ For $r=7$ $$\begin{align}P_{7+1}=\dfrac{8-7}{7+1} P_7\\ \implies P_8=\dfrac{1}{8} \cdot 8=1\end{align}$$ For $r=8$ $$\begin{align}P_{8+1}=\dfrac{8-8}{8+1} P_8\\ \implies P_8=0\end{align}$$ So, $0=P_9=P_{10}=...$, hence the Pascal sequence of $n=6$ is $1,8,28,56,70,56,28,8,1,0,0,0, \ldots$.

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